Length of Smallest Subsequence such that sum of elements is greater than equal to K

Given an array arr[] of size N and a number K, the task is to find the length of the smallest subsequence such that the sum of the subsequence is greater than or equal to number K.

Example:

Input: arr[] = {2, 3, 1, 5, 6, 3, 7, 9, 14, 10, 2, 5}, K = 35
Output: 4
Smallest subsequence with the sum greater than or equal to the given sum K is {7, 9, 14, 10}



Input: arr[] = {1, 2, 2, 2, 3, 4, 5, 4, 7, 6, 5, 12}, K = 70
Output:-1
Subsequence with sum greater than equal to the given sum is not possible.

Approach:

  • This problem can be solved with the help of priority queue
  • Traverse input array and insert every array element into priority queue.
  • Initialize variables that holds the sum of picked element from priority queue and the variable to get the count of picked element from priority queue to 0
  • Pop the elements out from the priority queue untill the priority queue is not empty
    1. Add the element in into the sum
    2. Increase the count because the element is picked to contribute into the total sum
    3. Check if the sum is greater than the given number K, If yes then stop checking and output the count.

Below is the implementation of above approach.

C++

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// C++ implementation to find length of smallest
// subsequence such that sum of elements
// is greater than equal to given number K
  
#include <bits/stdc++.h>
using namespace std;
  
// Function to find the smallest
// subsequence such that sum of elements
// is greater than equal to given number K
int lengthOfSmallestSubsequence(int K, vector<int> v)
{
    // Initialize priority queue
    priority_queue<int> pq;
  
    // Loop to insert all elements into
    // the priority queue
    for (int i = 0; i < v.size(); i++) {
        pq.push(v[i]);
    }
    int sum = 0, count = 0;
  
    // Loop to find the smallest
    // subsequence such that sum of elements
    // is greater than equal to given number K
    while (!pq.empty() && sum < K) {
        sum += pq.top();
        pq.pop();
        count++;
    }
    // If sum is less then K
    // then return -1 else return count.
    if (sum < K) {
        return -1;
    }
    return count;
}
  
// Driver code
int main()
{
  
    vector<int> v{ 2, 3, 1, 5,
                   6, 3, 7, 9,
                   14, 10, 2, 5 };
    int K = 35;
  
    cout << lengthOfSmallestSubsequence(K, v);
  
    return 0;
}

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Java

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// Java implementation to find length of smallest
// subsequence such that sum of elements
// is greater than equal to given number K
import java.util.*;
  
class GFG
{
  
// Function to find the smallest
// subsequence such that sum of elements
// is greater than equal to given number K
static int lengthOfSmallestSubsequence(int K, int []v)
{
    // Initialize priority queue
    Queue<Integer> pq = 
            new PriorityQueue<Integer>(Collections.reverseOrder());
  
    // Loop to insert all elements into
    // the priority queue
    for (int i = 0; i < v.length; i++) 
    {
        pq.add(v[i]);
    }
    int sum = 0, count = 0;
  
    // Loop to find the smallest
    // subsequence such that sum of elements
    // is greater than equal to given number K
    while (!pq.isEmpty() && sum < K)
    {
        sum += pq.peek();
        pq.remove();
        count++;
    }
      
    // If sum is less then K
    // then return -1 else return count.
    if (sum < K) 
    {
        return -1;
    }
    return count;
}
  
// Driver code
public static void main(String[] args)
{
    int []v = { 2, 3, 1, 5,
                6, 3, 7, 9,
                14, 10, 2, 5 };
    int K = 35;
    System.out.print(lengthOfSmallestSubsequence(K, v));
}
}
  
// This code is contributed by Rajput-Ji

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Python3

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# Python3 implementation to find length of smallest
# subsequence such that sum of elements
# is greater than equal to given number K
  
# Function to find the smallest
# subsequence such that sum of elements
# is greater than equal to given number K
def lengthOfSmallestSubsequence(K, v):
      
    # Initialize priority queue
    pq = []
  
    # Loop to insert all elements into
    # the priority queue
    for i in v:
        pq.append(i)
    pq.sort()
  
    sum = 0
    count = 0
  
    # Loop to find the smallest
    # subsequence such that sum of elements
    # is greater than equal to given number K
    while (len(pq) > 0 and sum < K):
        sum += pq[-1]
        del pq[-1]
        count += 1
      
    # If sum is less then K
    # then return -1 else return count.
    if (sum < K):
        return -1
    return count
  
# Driver code
v = [2, 3, 1, 5,
    6, 3, 7, 9,
    14, 10, 2, 5]
K = 35
  
print(lengthOfSmallestSubsequence(K, v))
  
# This code is contributed by mohit kumar 29

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Output:

4

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Improved By : mohit kumar 29, Rajput-Ji